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23-Chem-B1 Transport Phenomena · December 2016

Question 1 of 6: A1 — Turbulent pipe flow from a pressure gradient

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. EGBC 04-CHEM-B1 Transport Phenomena, December 2016, 3 hours, open-book. Six problems in three sections (A Fluid Mechanics, B Heat Transfer, C Mass Transfer); candidates attempt one from each section plus a fourth (four of six at 25 marks each, only the first four in the answer book are marked). All six problems are solved below as a complete study resource. Appendix A of the paper supplies the equations of change (continuity, Navier–Stokes, energy and species tables) in rectangular, cylindrical and spherical coordinates — these are quoted rather than re-derived.

Reference texts: R. B. Bird, W. E. Stewart & E. N. Lightfoot, Transport Phenomena (2nd ed., Wiley) — the equations of change and the differential momentum / energy / species balances; J. R. Welty, C. E. Wicks, R. E. Wilson & G. L. Rorrer, Fundamentals of Momentum, Heat and Mass Transfer (Wiley) — pipe friction, internal-flow heat transfer and film mass transfer; F. P. Incropera & D. P. DeWitt, Fundamentals of Heat and Mass Transfer (Wiley) — the log-mean-temperature-difference and effectiveness–NTU methods; R. H. Perry & D. W. Green, Perry’s Chemical Engineers’ Handbook (9th ed.) — transport properties and the Colebrook friction correlation.

Question 1: A1 — Turbulent pipe flow from a pressure gradient (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

QuantitySymbolValue
Internal diameter$D$$7\ \text{cm}=0.07\ \text{m}$
Pressure gradient$\Delta P/L$$125\ \text{Pa/m}$
Density$\rho$$997\ \text{kg/m}^3$
Viscosity$\mu$$8.57\times10^{-4}\ \text{Pa}\!\cdot\!\text{s}$
Relative roughness (part b)$\varepsilon/D$0.002

Find. (a) the volumetric flow rate $Q$ in the smooth pipe; (b) the average velocity when the pipe has relative roughness 0.002.

smooth pipe, D = 7 cm D high p low p flow direction, ΔP/L = 125 Pa/m
Fig. A1: Steady flow in a horizontal circular pipe driven by a constant axial pressure gradient. The wall shear balances the pressure force; the flow turns out to be firmly turbulent, so a friction-factor correlation (not Hagen–Poiseuille) sets the velocity.

Approach. The pressure gradient fixes the product $f\,v^2$ through the Darcy friction relation; test the laminar assumption first (it fails badly), then iterate a smooth-pipe (part a) and rough-pipe (part b) friction correlation with the Reynolds number until the velocity converges.

  1. Friction relation ties $\Delta P/L$ to $v$. For fully-developed pipe flow the Darcy–Weisbach law gives $\dfrac{\Delta P}{L}=f\,\dfrac{\rho v^2}{2D}$, so the pressure gradient fixes $$f\,v^2=\frac{2D}{\rho}\,\frac{\Delta P}{L}=\frac{2(0.07)(125)}{997}=1.755\times10^{-2}\ \text{m}^2/\text{s}^2 .$$
  2. Reject the laminar assumption. If the flow were laminar, Hagen–Poiseuille $Q=\pi(\Delta P/L)R^4/(8\mu)$ would give $v=Q/A\approx22\ \text{m/s}$ and $Re=\rho vD/\mu\approx1.8\times10^{6}$ — wildly turbulent, contradicting the assumption. The flow is therefore turbulent and a friction-factor iteration is required.
  3. Iterate the smooth-pipe correlation (part a). With $\varepsilon=0$ the Colebrook equation $\dfrac{1}{\sqrt f}=-2\log_{10}\!\Big(\dfrac{2.51}{Re\sqrt f}\Big)$ (equivalently Blasius $f=0.316\,Re^{-1/4}$) is solved together with $v=\sqrt{f\,v^2/f}$ and $Re=\rho vD/\mu$. Starting from a guess and iterating to convergence: $$\boxed{v\approx0.963\ \text{m/s},\quad Re\approx7.84\times10^{4},\quad f\approx0.0189.}$$ The Blasius formula gives $f=0.316(78400)^{-1/4}=0.0189$, confirming the Colebrook root.
  4. Flow rate (part a). $A=\dfrac{\pi D^2}{4}=\dfrac{\pi(0.07)^2}{4}=3.848\times10^{-3}\ \text{m}^2$, so $$Q=vA=0.963(3.848\times10^{-3})\;\Longrightarrow\;\boxed{Q\approx3.71\times10^{-3}\ \text{m}^3/\text{s}=3.71\ \text{L/s}.}$$
  5. Rough pipe (part b). Re-solve the same $f\,v^2$ constraint with the full Colebrook equation including $\varepsilon/D=0.002$: $\dfrac{1}{\sqrt f}=-2\log_{10}\!\Big(\dfrac{\varepsilon/D}{3.7}+\dfrac{2.51}{Re\sqrt f}\Big)$. Iterating gives $$\boxed{v\approx0.824\ \text{m/s},\quad Re\approx6.71\times10^{4},\quad f\approx0.0258.}$$
  6. Interpret the roughness penalty. Roughness raises the friction factor from 0.019 to 0.026, so for the same pressure gradient the pipe carries a lower velocity (0.824 vs 0.963 m/s, about 14 % less) and a correspondingly lower flow rate $Q\approx3.17\times10^{-3}\ \text{m}^3/\text{s}$.
QuantityResult
(a) Smooth-pipe velocity$v\approx0.963\ \text{m/s}$ ($Re\approx7.8\times10^4$, turbulent)
(a) Volumetric flow rate$Q\approx3.71\times10^{-3}\ \text{m}^3/\text{s}\ (3.71\ \text{L/s})$
(b) Average velocity, $\varepsilon/D=0.002$$v\approx0.824\ \text{m/s}$ ($f\approx0.0258$)
Check — laminar vs turbulent

Always check the regime before choosing a friction law. Here the pressure gradient is far too large for laminar flow: assuming laminar and applying Hagen–Poiseuille returns $Re\sim10^6$, which is self-contradictory. The turbulent iteration converges to $Re\approx7.8\times10^4$, comfortably above the $Re\approx4000$ transition, so the smooth Colebrook / Blasius correlations apply.

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